【文档说明】广西钦州市2019-2020学年高二下学期期末教学质量监测数学（文）试题扫描版含答案.doc，共(7)页，2.614 MB，由小赞的店铺上传

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钦州市2020年春季学期教学质量监测参考将答案高二数学（文科）一、选择题答案：（每小题5分，共60分）题号123456789101112答案BBDDCCDABBCD二、填空题答案：（每小题5分，共20分）13.(1,2)；14．5；15．2；16．8月份利润最低（或“3月份和10月份利润最

高”等其他正确结论）三、解答题：本大题共6题，共70分．解答应写出文字说明、证明过程或演算步骤．17.证明：要证511313只须证22511313·····························

···········································2分只须证525511323913·····················································

·················.4分只须证5539··························································································

·····6分只须证5539·································································································8分因为5539

成立所以511313·································································10分18．解：（1）所有感染病毒的小白鼠共有50只，其中注射疫苗的共有y只，∴1505yP·············

································································································2分∴10,501040,yx402060,301040,mn

·····························4分∴22列联表如下：未感染病毒感染病毒总计未注射疫苗204060注射疫苗301040总计5050100·································

····································································6分（2）∵22100(20103040)1005016.6675050604063K

····································10分∵16.6676.635∴有99%把握认为注射此种疫苗对预防新型冠状病毒有效.························12分19．解：（1）（

1）∵cosx，siny，·························································2分∴1C的极坐标方程为sin2．···························

····································4分（2）∵直线3C的极坐标方程为6πR∴124sin6π，····························································

········································7分24sin()23,66··································································

························10分∴21423MN·····································································

··············12分20.解：∵()|2|1fxxx，()7fx∴①当1x时，原不等式可化为217,x解得[3,1]x；··························3分②当12x时

，原不等式可化为37,解得(1,2)x；···································6分③当2x时，不等式可化为217,x解得[2,4]x；·················

·························9分综上，原不等式的解集为[3,4]···········································································12分21.解：（1）∵

32cos,2sinxy∴2222(3)2cos2sin4xy∴曲线C的直角坐标方程22(3)4xy····················································

···········4分（2）∵直线l的参数方程为222,22xtyt（t为参数）∴直线l与x轴交于点(2,0)P················································································

6分将222,22xtyt代入圆C的方程22(3)4xy，整理得2230,tt122,tt12·3tt··························································

···········································8分由参数的几何意义得1212PMPNtttt21212()4tttt2(2)4(3)14.∴14MN

PMPN·········································································12分22.解：（1）由表中数据得，23453.5

4x，26394954424y，∴41421ˆ9.4iiiiixxyybxx，429.43.59.1aybx，∴回归方程为9.49.1yx．··········

·········································································6分（2）年资金投入量为7.5万元时，9.47.59.179.6y（万元）；·

···················8分（3）解9.49.1100x，······························································································10分得90.99.4x

∵90.99.679.4∴若年市场销售额超过100万，那么年资金投入量至少要9.67万元.··········12分